Lecture 14
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Lecture 14 – Thursday 2/24/2011
 Pseudo Steady State Hypothesis (PSSH)
 Net Rate of Reaction of Active Intermediates is Zero
 Hall of Fame Reaction: 2NO +O2 2NO2
 Introduction to Enzyme Kinetics
 Begin Non-Isothermal Reactor Design
2
An active intermediate is a molecule that is in a highly
energetic and reactive state It is short lived as it disappears
virtually as fast as it is formed. That is, the net rate of
reaction of an active intermediate, A*, is zero.
The assumption that the net rate of reaction is zero is
called the Pseudo Steady State Hypothesis (PSSH)
3
4
The rate law for the reaction
A  BC
is found from experiment to be
 rA 
kC
2
A
1  k C A
How did this rate law come about? Suggest a mechanism
consistent with the rate law.
5
For reactions with active intermediates, the reaction
coordinated now has trough in it and the active intermediate,
A*, sits in this trough
6
7
1
A  A  A *  A
r1 A*   k 1C A
2 
A *  A  A  A
r2 A *   k 2 C A * C A
3 
A *  B  C
r3 A *   k 3 C A *
k1
k2
k3
2
k3 is defined w.r.t. A*
Reaction (1)
r1 A*  k 1C
Reaction (2)
r2 A *   k 2 C A C A *
(2)
Reaction (3)
r3 A *   k 3 C A *
(3)
2
A
(1)
But C*A cannot be measured since it is so small
r1 A   r1 A* , r3 B   r3 A*
8
Rate of Formation of Product
rB  r3 B   r3 A*  k 3 C A*
(4)
Pseudo Steady State Hypothesis r*A = 0
r A *  r1 A *  r2 A *  r3 A *
(5)
 k 1C  k 2 C A C A *  k 3 C A *  0
(6)
rA* 

2
A
Solving for C A *
C A* 
9
k 1C
2
A
k 3  k 2C A
Substituting for C A* in Equation (4) the rate of formation
of B is
2
k1 k 3C A
(8)
rB 
k 3  k 2C A

Relative rates overall
A  BC
rA
1

rB
1
2
rA   rB  
k1 k 3C A
(9)
k 3  k 2C A
10
For high concentrations of A, we can neglect k 3 with regard to
k 2 C A , i.e.,
k 2 C A  k 3
and the rate law becomes
rA  
k1k 3
k2
(10)
C A   kC A
Apparent first order.
11
For low concentrations of A, we can neglect
to k3, i.e.,
k 2C A
with regard
k 3  k 2 C A
and the rate law becomes
rA  
k 3 k1
k3
C A   k 1C A
2
2
(11)
Apparent second order.
Dividing by k3 and letting k’=k2/k3 and k=k1 we have the rate
law we were asked to derive
2
 rA  
kC A
1  k C A
(12)
12
Why so many Reactions Follow Elementary Rate Laws
 r A  kC A
What about
 rA 
kC A
( I  Inert )
1  k `C I
(1)
1
A  I 
A*I
k
2
( 2 ) A *  I 
AI
k
3
( 3 ) A 
BC
k
- rA 
13
k1k 3C A
k 2  k 2C I
Hall of Fame Reaction
The reaction
2NO +O2 2NO2
has an elementary rate law
2
rNO 2  kC NO C O 2
However… Look what happens to the rate as the temperature is
increased.
-rNO2
14
T
Why does the rate law decrease with increasing temperature?
:
NO  O 2  NO
k1
*
3
NO  NO  O 2
(2)
NO  NO  2 NO 2
(3)
*
3
*
3
k2
k3
Write Rate of formation of Product rNO2
15
(1)
Note: k3 is defined w.r.t. NO2
NO3*
Assume that all reactions are elementary reactions, such
that:
r1NO3 = k1CNOCO2 = k1 [ NO][O2 ]
r2 NO* = -k2CNO* = k2 éë NO ùû
3
3
*
3
r3NO* = -k3CNO* CNO
3
3
*
é
= k3 ë NO3 ùû [ NO]
rNO = 2 é -r3NO* ù
ë
2
3 û
16
The net reaction rate for NO3* is the sum of the individual
reaction rates for NO3*:
r1NO* = r1 Þ r1NO* = k1[ NO] [O2 ]
3
3
-r2NO* = r2 Þ r2NO * = -k2 [ NO3* ]
3
3
r3
1
-r3NO = Þ r3NO = - k3 [ NO3* ] [ NO]
2
2
*
3
*
3
rNO* = r1NO* + r2NO* + r3NO*
3
17
rNO*
3
3
3
3
1
= k1[ NO] [O2 ] - k2 [ NO ] - k3 [ NO3* ] [ NO]
2
*
3
The PSSH assumes that the net rate of species A* (in this
case NO3*) is zero.
rNO *  0
3
1
*
é
ù
é
0 = k1 [ NO] [O2 ] - k2 ë NO û - k3 ë NO3 ùû [ NO ]
2
0 = k1 [ NO] [O2 ] - éë NO3* ùû ( k2 + [ NO])
*
3
k1 [ NO ][O2 ]
éë NO ùû =
k2 + k3 [ NO ]
*
3
18
NO 
*
3
k 1  NO O 2 
k2 
k3
2
 NO 
rNO2 = -2r3NO* = 2 éë NO3* ùû [ NO]
3
k1k3 [ NO ] [O2 ]
=
k3
k2 + [NO]
2
2
rNO2
19
k2 >> k3 [ NO]
rNO 2 
k1k 3
k2
 NO  O 2  
2
A1 A3
e
RT
A2
E 2   E1  E 3 
20
E 2   E1  E 3 
The result shows why the rate
decreases as temperature increases.
 NO  O 2 
2
-rNO2
T
21